Question 1206762
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The responses from the other tutors show solutions using the standard method, involving the fraction of the work performed by each worker in 1 day.<br>
I personally find solving this kind of problem easier using a different method.<br>
Consider workers A and B first.  A and B together can do the work in 12 days; B alone can do it in 18 days.<br>
Consider the least common multiple of those two times, which is 36 days.  In 36 days, the two together could do the job three times and B alone could do the job twice.  That means in those 36 days A could do the job once.  So A alone would take 36 days to do the work.<br>
Now we can use the same process to find how long it would take workers A and C to do the job together.  C can do the work alone in 20 days; A can do it alone in 36 days.  The least common multiple of those two times is 180 days.  In 180 days A could do the job 5 times and C could do the job 9 times.  So together the two can do the job 14 times in 180 days, which means the number of days they would take together to do the one job is 180/14 = 90/7 = 12 6/7 days.<br>
ANSWER: 12 6/7 days<br>